Finding the force exerted by one plate on another

[mr_coffee]

Hello everyone, I'm stuck on a mutliple choice question: A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 5 10-6 C then the force exerted by one plate on the other has a magnitude of about:
1 E4 N
9 E5 N
5 N
9 N
0

Here is what i did:
I know the E-field right outside a conductor is E = (Q/A)/Eo <--permitivity. So i found E = (5e-6/.3)/8.85e-12 = 1883239 N/C; Then i knew F = QE, so I said, F = (5e-6)(1883239) = 9.4N, which was wrong, then i thought, hm..maybe i messed up a unit somehow, so i tried 9e5N, also wrong, then i thought, well if they are both positive, that means the net force is going to be 0, because ulike charges repel, same charge, 0 net force. also wrong. What did i do wrong?

I also ran into odd problem, I'm wondering if the homework problem is wrong...
A 20 F capacitor is charged to 200 V. Its stored energy is:
4000 J
0.1 J
2000 J
0.4 J
4 J

Easy enough, U = .5*CV^2;
U = .5(20)(200)^2 = 400000J;
The answer is .4J though?

 

[quantumdude]

Sorry for the late reply. If you're still interested, then here are my comments.

I know the E-field right outside a conductor is E = (Q/A)/Eo <--permitivity.

Check that expression, because I don't think that you can use it when dealing with parallel plate capacitors.

See the following websites:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html
http://dept.physics.upenn.edu/~uglabs/lab_manual/electric_forces.pdf

 

[quicknote]

Hello,

I can provide some clarification on the problem you are facing. First of all, your calculation for the electric field is correct. However, the mistake lies in your calculation for the force exerted by one plate on the other. The correct equation for the force between two charged plates is F = (Q1Q2)/4πε0d^2, where Q1 and Q2 are the charges on the two plates and d is the distance between them. In this case, the force exerted by one plate on the other would be F = (5e-6)(5e-6)/4π(8.85e-12)(0.1e-3)^2 = 9N. Therefore, the correct answer to the multiple-choice question would be 9N.

As for the second problem, you are correct in your calculation for the stored energy. The answer of 0.4J is incorrect and it seems like there may be a mistake in the problem itself. I would recommend double-checking the units and values given in the problem to see if there is an error.

I hope this helps! Keep up the good work in your studies. As a scientist, it is important to always double-check our calculations and question any discrepancies that we encounter.